Factoring drag into artillery flight, differential equation

Imagine separating a parabola into its different x and y components and linking them by the component of time.

Scenario:
Given:
Gravity
No Drag
The velocity of the launched object is not sufficient to escape the gravitational pull.
Ignoring:
The change of gravity as the distance between the centre of the two masses increases (projectile and planet constitute as the two masses)
An object is launched on a flat plain, at an angle of Θ, with a velocity of Ω.
There are two components to this parabola; the constant x velocity, the changing y velocity (due to gravity)
The x velocity is
X=(Ω(90-Θ))⁄90
This equation splits the initial velocity into x and y values. Imagine if you launched an object strait up, there is no x velocity. If an object is launched at a gradient of tan⁡45 (tan is equivalent to rise over run) then the initial velocity will be evenly divided between x and y.
Note x is constant
The x distance:
Xdist=time(( Ω(90-Θ))⁄(90 ))
The y velocity:
The y velocity is constantly changing due to the nature of gravity.
The initial y velocity is
Y=(Ω(Θ))⁄90
And now gravity is taken into account

Velocity of falling object:
V= Velocity
A = Acceleration constant
T= Time
V=AT
Gravity’s pull is a negative addition to the y velocity.
Therefore the overall y velocity can be summarised as:
Yvel=(Ω(Θ))⁄(90 )-AT
We don’t want velocities; we need distances to find the equation of the parabola.

A
To find the equation for this graph one would find the distance equation,
V=at
If this is integrated (D=distance)
D=1/2 at^2
Therefore the overall equation for the y height is:
YHeight= -1/2 at^2+ t(Ω(Θ)⁄(90 ))
Xdist=time(( Ω(90-Θ))⁄(90 ))
Given that the x distance is related to the y distance by time we can combine the equations to get the parabola that is formed. Or alternately 3 time points can be substituted into the above x and y equations along with all the other information required, then one can solve simultaneously for the equation. The constant (c) in the general form of the quadratic equation y=ax^2+bx+c can be found, it is equal to the height of launch.
The next step is to factorise in drag, into the model of the flight path of the projectile, unfortunately at this point parabolas cannot model the flight path of the object.
The quadratic drag equation
Circia 1900’s, Lord Rayleigh
Fd=-1/2 pv^2 ACdv
Where
Fd= force of drag, p = the density of the fluid, v=velocity, A= reference area,

Now
Consider.
A force instantaneous affair, given this if a relationship to time can be realised from this equation, and it can then be integrated, the sum total of the entire negative vector up until a point can calculate giving the velocity at the corresponding time point.
One of the properties of air resistance is that due to its non linear nature the drag must be calculated on the total velocity, and then separated into its x and y components.
The velocity of the object is:
V=( Ω(90-Θ))⁄(90 ))*p +(Ω(Θ))⁄(90 )-AT*p
V= velocity
P=launch velocity

this is as far as i have gone.

How can drag be added to the different x and y components with a link to time?

note: all this has been done with year 11 maths and i could be completely wrong, this is not homework, this is just out of interest. how can the path of a rocket which has an non linear (change in gravity and air resistance) accelerating velocity. non linear changing angle of flight respective to (center of gravity, velocity, angle of launch, and much more ) be modeled.

Pretty much everything you have written is wrong. "Friction" is a force and so affects acceleration, not velocity directly. Also the trajectory will NOT be a parabola. I hate to have to tell you this but your teacher is correct (don't you just hate when that happens), you will have a second order differential equation. "Acceleration" is the derivative (instantaneous rate of change) of velocity and velocity itself is the derivative of the position function.

For reasonably short distance, close to the earth's surface, the acceleration due to gravity is a constant, g, so if we model friction as proportional to the speed, we get the differential equations,
[tex]m\frac{dv}{dt}= -g- kv[/tex]
and
[tex]m\frac{dx}{dt}= v[/tex]
which can be combined to
[tex]m\frac{d^2v}{dt^2}= -g- k\frac{dx}{dt}[tex]
That equation will, typically, give exponentials as solutions.

thanks,
i thought that if retardation from drag was ignored,
if an object was launched with a initial velocity, constant x velocity, and accelerating y velocity (gravity) it would follow the shape of a parabola.

thanks,
i thought that if retardation from drag was ignored,
if an object was launched with a initial velocity, constant x velocity, and accelerating y velocity (gravity) it would follow the shape of a parabola.

Hi 3physqs. It's actually only a parabola if you take the Earth to be flat. Of course parabolic is an excellent approximation for a short near Earth trajectory.

The reason for the discrepancy is that the gravitational force is (ideally) radially directed (towards the center of the Earth), so both it's magnetude and direction change thoughout the flight. To get a parabola you need the gravitational force to be unchanging (both magnetude and direction) throughout the entire flight.